Electromagnetic Interactions
Introduction to Elementary Particle Physics
Diego BettoniAnno Accademico 2010-2011
Outline
• Electron-nucleus scattering • Rutherford formula
– Mott cross-section• Electron-nucleon scattering • Rosenbluth formula
– Hadron form factors• The process e+e- +-
• Bhabha scattering e+e- e+e-
• Magnetic moments of leptons– Measurement of the muon g-2.
e-nucleus Scattering(neglecting spin)
The transition probability (per unit time) is given by the Fermi golden rule:
For a potential V(r) the matrix elementis given by the volume integral:
i = wave function of the incoming electronf = wave function of the scattered electronThe Born approximation assumesthe perturbation to be weak; we canrepresent i and f as plane waves
where k0=p0/ħ e k=p/ħ are the initial and final propagation vectors, respectively.
0 0_
,p_
W,p’
E,p_
-
E
e
θe A
A
-
fifMW 22
drVM ifif )(*
rdrVeM rkkiif
3)( )(0
ff dE
dphdp3
2
fif dE
dphdpMdW 3
222
vdWd
22
4221
iff
MdEdp
vp
dd
Ef = total energy in the final state
v =velocity of the incident beam
Let us now consider the nuclear recoil. We assume that both incident andscattered electrons are extreme relativistic (v) and we set ħ=c=1.
p0 = k0 = E0 p = k = E From energy conservation:
Ei = p0+M = p+W = Ef
initial nucleus finalenergy mass energy
From momentum conservation:
ppp 0
2
022
0
220
22
cos2 Mppppp
Mppp
MppWpE f
00 cos pp
MW
pEW
dEdp
ff
2sin21
1
)cos1(1
12000
Mp
Mpp
p
23
0
22 )(
41
rdrVe
pp
MWp
dd rqi
ppq 0 momentum transfer
0
00
0
0
20
220
0
cos
cos
cos
cos22cos221
11
pp
MW
pMpM
MW
pEW
ppWW
Mpppppp
dpdEdE
dp
f
ff
)cos1(1
1cos
0cos
2:
cos2222
cos2
cos2
0000
02000
20
20
22000
2220
20
220
20
20
2200
MppMp
Mpp
pp
ppM
pp
pM
p
MppppMpppMppMp
MpppppMp
MpppppMp
R
r
Rrs
RdRZedq 3)(
e
Let us now calculate the matrix element, taking for V(r) the coulomb interaction.We represent the nucleus by a sphere of charge density )(RZe
1)( 3 RdR
RrRdRZe
sedqdV
4)(
432
Rr
RdRZerV
4)()(
32
)(cos2)(4
)(4
4)(
2cos
32
)(33
2
233
dsdss
eeRRdZe
RrReeRdrdZe
RrRZeRderdM
iqsRqi
RqiRrqi
rqiif
polar anglebetween s and q
Let us define the Nuclear Form Factor RdeRqF Rqi 3)()( If (R) is spherically symmetric (R) = (R)
22
32
611
)(2
)(1)(
Rq
RdRRqRqiqF 22 qq
)( 2qFF
The matrix element becomes:
0
22
1
1
cos
0
22
)(2
)(cos)(2
dsiqs
eesqFZe
desdsqFZeM
iqsiqs
iqsif
This integral diverges.We modify V(r) by a factor which takes into account the screening of thenucleus by atomic electrons.Since a atomic dimensions >> R nuclear dimension, we can write:
are /
aRee asar //
And the matrix element becomes:
0
/22
)()(2
dseeeqFiq
ZeM iqsiqsasif
22
22
)1()1(
12
1
11
11
11
aq
iq
aq
iqa
iqa
iqa
iqa
dsedseiq
asiq
as
22
22
1)(
aq
qFZeMif
a 10-8 cm = 10-10 m 5105 GeV-1
1/a 0.210-5 GeV = 2 keVq2 >> (1/a)2
)( 22
2
qFqZeMif
22
0
24
222 )(1
44 qF
pp
MWp
qeZ
dd
Let us assume 1MW
(Non relativistic nuclear recoil)
)1(0
00 pppppqp
2sin4cos22)( 22
020
20
20
2 pppppq
22
420
22222
044
0
222 )(
2sin4
)(
2sin16
14
4 qFp
ZqFpp
eZdd
]);(cos2);(cos2[ 220
20
2 dqp
ddpdqdd
For an effectively pointlike nucleus, i.e. low values of q2, F(q2) 1
4
22
2
420
22
42
sin4
qZ
dqd
p
Zdd
Rutherford
cross section
Four-momentum Transfer
)0,(),(
0
000
MPpEP
Initial state
incident electron
nucleus at rest in the laboratory
),(),(pWP
pEP
Final state
scattered electron
recoiling nucleus
2sin4)cos1(2
cos222
22
200
002
022
0022
0
20
20
20
2
pppp
ppEEm
ppppEEEE
ppEEPPq
Em
q2 < 0 spacelike (scattering processes)q2 > 0 timelike (annihilation, e.g. e+e- +-)
Considering the four-momentum transfer to the nucleus:
MKMWMpWMPPq 222)()( 22220
2
where K = W-M = kinetic energy of the nucleus
Thus: 2
2
21
Mq
MW
The nucleus recoils coherently for q2 << 2M2
1MW
The nuclear form factor:
dRRqR
qRR
dRRiqR
eeR
dRdReRqFiqRiqR
iqR
2
2
2cos2
4sin)(
2)(
)(cos2)()(
Typical nuclear radius is R a few fm. For example, if R = 4 fm, qR = 1, q=1/R
MeVfm
fmMeVRcqc 50
4197
Therefore if q << 50 MeV/c, qR 0 and F(q2) 1
4
2
24
qdqd
21q
e e
p p
Electron Spin
For a relativistic fermion the spin vector is aligned with the momentum vector
if defines the z axis.The helicity H is defines as:
p
01 yxz p
1
ppH
H=+1 right-handed RH= -1 left-handed L
In electromagnetic interactions helicity is conserved. L L R R R L
e- e- e- e- e+ e-
Jz = 1 Transverse photon
In the relativistic limit fermion and antifermion have opposite helicities.
L
e
L
L
L
allowed forbidden
Jz =-½
Jz =-½
Helicity amplitudes
jmmd
2sin
21
21
2cos
21
21
2cos
21
21
21
21
21
21
21
21
21
21
21
,
,
,
dJJ
dJJ
dJJ
zz
zz
zz
2sin21
1
2sin4
2cos
20420
222
Mpp
Z
dd
Mott
Mottcross section
the factor p/p0takes into accountthe nuclear recoil.
e-N Scattering
Let us now consider also the spin of the target. In the scattering of electrons byhypothetical pointlike protons there will be a magnetic and an electrical interaction.
2
sin22
cos 22
22
Mq
dd
dd
RutherfordDirac
2sin21
1
2sin4 2042
0
22
Mpp
Z
electrical(non spin-flip)
magnetic(spin-flip)
If nucleons were pointlike this would be the cross section. However p and n are notpointlike, as shown by their anomalous magnetic moments:
(Dirac) (experimental)p eħ/2mc = 1 n.m. +2.79 n.m.n 0 -1.91 n.m.
Nucleons have anextended structure
Proton Structure
p0 p
k k
jJ
xkki
xppi
ekukueJ
epupuej
)(
)(0
)()(
)()( 0
most general 4-vector whichcan be constructed from k, k, q and the Dirac matrices .
There are only two independent terms, and iq,and their coefficients are functions of q2.
qiqF
MqF 2
22
1 2
F1 and F2 are the Form Factors of the Proton
As q2 0 we see a particle of charge and magnetic momente Mc
e2
1
F1(0) =1 F2(0)=1 protonF1(0) =0 F2(0)=1 neutron
2
sin22
cos4
22212
222
22
222
1 FF
MqF
MqF
dd
dd
Rutherford
Sachs Form Factors
21
22
2
1 4FFG
FMqFG
M
E
electric Form Factor
magnetic Form Factor
2
sin22
cos1
22222
MME
Rutherford
GGGdd
dd
2
tan21
2222
MME
Mott
GGGdd
dd
Rosenbluth Formula
2
2
4Mq
22 qGGqGG MMEE 91.1079.20
0010
nM
pM
nE
pE
GGGG
2
tan222
qBqA
dd
dd
Mott
RosenbluthPlot
The experimental determination of the nucleon form factors in the spacelike region(q2 < 0) is carried out by directing e- beams of energy between 400 MeV and16 GeV at a hydrogen target (for the proton) or deuterium (for the neutron).
dedepepe
For the neutron:
correzionedifattoriepdded
dden
dd
)()(
Scaling laws for the form factors:
02
222
2
qG
qGqGqGqG
nE
n
nM
p
pMp
E
Nucleon Spacelike Form Factors
p
pMG
n
nMG
pEG
Proton Form Factors at High q2
Dipole formula: 2
2
2
2
1
1
VMq
qG22 )84.0( GeVMV
RMVeR 0)(
2
0
30
0
320
2 12VRM
RM
MRde
RdReR
V
V
fmM
RV
80.0122
The dipole form corresponds to an exponentialcharge distribution
with an rms radius
For the proton:
• Different charge and magnetization distributions
• Quark angular momentum contribution?
Rosenbluth
polarization
Linear deviationfrom dipoleGE≠GM
Form Factor Measurements Using Polarization
Timelike Proton Form Factors
They can be measured via the processes e+e- pp, or pp e+e-. For the latter in the CMS frame:
*22
2*22
222
* cos14
cos12cos
E
pM G
sm
Gxs
cd
d
p(E,p) pe+
e-*
22
22
*cos
*cos
2
0
42
)*(cosmax
max
Ep
Mp
GBsm
GAs
dddd
s = CM Energy
One measures thetotal cross section .
Proton Magnetic Form Factor
222 ln ss
CG
p
M
The dashed line is afit to the PQCD prediction
The expected Q2 behaviouris reached quite early, however ...... there is still a factor of 2between timelike and spacelike.
Timelike Form Factor of the
ee e+ beam with energies100, 125, 150, 175 GeV
Spacelike Form Factor of the
ee 300 GeV - beam
fmr
fmr
66.0
008.0439.0 22
e+e- +-
e+ e-
+
-
s
ee3
4 2
s = 4E1E2 if s >> me2, m
2
21q
e
e
•Each vertex gives a contribution () to the matrix element. The cross sectionis therefore proportional to 2. 2.
•For a timelike process q2=s, hence the propagator
• has dimensions of (length)2, i.e. (energy)-2. If s >> me2, m
2
s is the only energy scale in the process:•The factor (4/3) comes from integration over solid angle and averaging over spins.
sq11
2
s1
Angular Distribution for e+e- +-
2cos1d
d
11 zJJ
Because electromagnetic interactions conserve parity, both Jz=+1 and Jz=-1 occurwith equal probability. The amplitude for emitting the + at angle to the e+
starting from a JZ=+1 state is given by:
cos1211
1,1 dIf the initial state has JZ=-1 we have to replace by - hence the amplitude becomes:
cos121
Squaring and adding the amplitudes of these two orthogonal states we obtain:
222 cos1cos1cos1 d
d
The process e+e- +- (ore+e- +-) is not purelyelectromagnetic: there is aweak contribution, due toZ0 exchange.
e
e
e
e
0Z
G G
ceinterferenddweak
ddQED
dd
dd
)(
s
2 sG2 G
The asymmetry arises from the interference term, the effect is of the order of10 % for s = 1000 GeV2.
e+e- e+e-
Bhabha Scattering
The dimensional arguments used for e+e- +- apply equally well to BhaBha scattering to predict a 1/s dependence for the total cross section.The angular distribution is however more complex, because two diagrams contribute:
The first diagram dominates at small angles. In this region the cross sectionis large and is used to monitor the luminosity in e+e- colliders.
Lepton Magnetic Moments
According to the Dirac theory a pointlike fermion possesses a magnetic momentequal to the Bohr magneton . If e and m are the lepton charge and mass:
In general the magnetic moment is related to the spin vector s by:
Where g is called the Landé factor and gB is the gyromagnetic ratio.For electron and muon |s|=½ and the Dirac theory predicts g=2.The actual g-values have been measured experimentally with great precisionand have been found to differ by a small amount (0.2 %) from the value 2.The Dirac picture of a structureless, point particle is not exact for the electronand muon.
me
B 2
sg B
The magnetic moment of a charged particle depends on the spatial distributions of charge and mass (e/m ratio). For a spin ½ a value g≠2 argues that processes are taking place which distort the relative chargeand mass distributions. For example, for the proton g=5.59, due to itsinternal structure.The electron, the , the consist of a bare, pointlike object surrounded bya cloud of virtual which are continually being emitted and reabsorbed.These carry part of the mass energy of the lepton, and hence thee/m ratio (and thus the magnetic moment) changes.In terms of QED:
The Landé can be written as a perturbation series in (/).To lowest order:
meg
22
At the next order:
1
22
meg
We can define the anomaly:2
2 ga
12
32
10)281159652140(
19.132848.05.02
2
QED
ee
ga
12
32
10)28181165847008(
45.2476578.05.02
2
QEDga
For the the measured value differs by 9 standard deviations from the QED calculation. This arises from the fact that for the there are furthercorrections to a due to the strong and weak interactions.Hadrons do not couple directly to the , but they can couple to the virtual photon. We therefore expect hadronic contributions to vacuum polarization of the kind:
This contribution, which would be small due to the high mass, gets amplifiedby resonances in the system (vector resonances).
-+
The strong contribution to the anomalycan be computed starting from the measurable cross section
e+e- hadronsvia dispersion relations the two diagramscan be related one to the other.
dss
adronieemfortea
03
2
12)(
Weak contributions
Measurement of the Muon (g-2)
Consider a longitudinally polarized charged particle moving in a staticmagnetic field B. The particle momentum rotates at the cyclotron frequency:
The spin precesses at the frequency:
If g=2, i.e. a=0, s=c and the particle will maintain its longitudinal polarization. If however g>2 (a>0), s>c, spin precesses faster than momentum. Therotation frequency a of the spin with respect to the momentum is given by:
mceB
c
mceBa
mceBgs 1
2
mceBacsa
The measurement principle of a is the following: muons are kept turning in a knownmagnetic field B, the angle between the spin and the direction of motion is measuredas a function of time and from this the value of a can be determined.Since a ≈ 1/800, the muon must make roughly 800 turns in the field for the spin tomake 801 and the polarization to change gradually through 2.
The muons take roughtly 2000 turns in the field. The field gradientdisplaces the orbit to the right. At the end a very large gradient is usedto eject the muons, which are then stopped in the polarization analyzer.
20 1 byayBBz
stcMeVpTB 2.2/906.1
x
y
Polarization analyzer. When a muon stopsin the liquid methylene iodide (E) a pulse ofcurrent in coil G is used to flip the spin through900. Backward or forward decay electronsare detected in the counter telescopes 66 and 77. The asymmetry in counting rates as a function of time is given by:
sAccccA sin0
BtmceaAA sin0
Using this method the anomaly awas measured with an accuracy of 0.4 %
61051162 a
Muon Storage RingIn order to improve the measurementaccuracy it was necessary to increasethe number of (g-2) cycles, either byincreasing the field B or by lengtheningthe storage time. The usage of an electric field for thevertical focussing allowed the use ofa uniform magnetic field.For the precession of spin in combinedelectric and magnetic fields:
EaBa
mce
a
11
2
For =29.3 (magic ) the coefficientof the second term vanishes and theprecession is again a.
This method allowed to considerably increase the accuracy in the measurement of a.
91091165924 a
aSM [e+e– ] = (11 659 182.8 ± 6.3had ± 3.5LBL ± 0.3QED+EW) 10 –10
BNL E821 (2004) : aexp = (11 659 208.0 5.8) 10 10
10102.92.25 SMexp aa
Present Situation for a
The discrepancy between theory and experiment is 2.7 standard deviations.